Since $f^{\prime}(x)>0$

$\Rightarrow f^{\prime}(x)$ is always increasing

$\mathrm{g}^{\prime}(\mathrm{x})=2 \mathrm{f}^{\prime}\left(2 \mathrm{x}^{3}-3 \mathrm{x}^{2}\right) \times\left(6 \mathrm{x}^{2}-6 \mathrm{x}\right)+\mathrm{f}^{\prime}\left(6 \mathrm{x}^{2}-4 \mathrm{x}^{3}-3\right)\left(12 \mathrm{x}-12 \mathrm{x}^{2}\right)$

$=12\left(\mathrm{x}^{2}-\mathrm{x}\right)\left(\mathrm{f}^{\prime}\left(2 \mathrm{x}^{3}-3 \mathrm{x}^{2}\right)-\mathrm{f}^{\prime}\left(6 \mathrm{x}^{2}-4 \mathrm{x}^{3}-3\right)\right)$

$\left.=12 \mathrm{x}(\mathrm{x}-1)]\left[\mathrm{f}^{\prime}\left(2 \mathrm{x}^{3}-3 \mathrm{x}^{2}\right)-\mathrm{f}^{\prime}\left(6 \mathrm{x}^{2}-4 \mathrm{x}^{3}-3\right)\right)\right]$

For increasing $g^{\prime}(x)>0$

Case – $I$ $\quad x<0$ or $x>1$

$\Rightarrow f\left(2 x^{3}-3 x^{2}\right)>f^{\prime}\left(6 x^{2}-4 x^{3}-3\right)$

$\Rightarrow 2 x^{3}-3 x^{2}>6 x^{2}-4 x^{3}-3$

$\left(\because f^{\prime}(x) \text { is increasing }\right)$

$\Rightarrow(x-1)^{2}\left(x+\frac{1}{2}\right)>0 \Rightarrow x>-\frac{1}{2}$

$\therefore x \in\left(-\frac{1}{2}, 0\right) \cup(1, \infty)$

Case $II$ : If $0$

$f^{\prime}\left(2 x^{3}-3 x^{2}\right)$

$(x-1)^{2}\left(x+\frac{1}{2}\right)<0$

$\Rightarrow x<-\frac{1}{2},$ so there is no solution

$x<-\frac{1}{2}$

$\Rightarrow$ Hence the values are $x \in\left(-\frac{1}{2}, 0\right) \cup(1, \infty)$